解决哲学家进餐问题可采用如下方法：a.仅当哲学家的左、右两只筷子均可用时，才允许他拿起筷子进餐；b.至多只允许有4位哲学家同时去拿左边的筷子，最终能保证至少有一位哲学家能够进餐；c.规定奇数号哲学家先拿起他左手的筷子，然后再拿起他右手的筷子，而偶数号哲学家则先拿起他右手的筷子，然后再拿起他左手的筷子。方法a在示例程序中给出，请用方法b和c写出不会发生死锁的哲学家进餐程序。

方法b：

import threading

class Philosopher(threading.Thread):
    def __init__(self, left_chopstick, right_chopstick, semaphore):
        threading.Thread.__init__(self)
        self.left_chopstick = left_chopstick
        self.right_chopstick = right_chopstick
        self.semaphore = semaphore

    def run(self):
        while True:
            # 拿起左边的筷子，最多只有4个哲学家同时拿左边的筷子
            self.semaphore.acquire()
            self.left_chopstick.acquire()
            # 拿起右边的筷子
            self.right_chopstick.acquire()

            # 进餐
            print(f"{threading.current_thread().name} is eating...")

            # 放下右边的筷子
            self.right_chopstick.release()
            # 放下左边的筷子，同时释放semaphore，允许其他哲学家拿起左边的筷子
            self.left_chopstick.release()
            self.semaphore.release()

if __name__ == "__main__":
    chopsticks = [threading.Lock() for _ in range(5)]
    semaphore = threading.Semaphore(4)

    philosophers = []
    for i in range(5):
        philosopher = Philosopher(chopsticks[i], chopsticks[(i+1)%5], semaphore)
        philosopher.name = f"Philosopher {i}"
        philosophers.append(philosopher)

    for philosopher in philosophers:
        philosopher.start()

方法c：

import threading

class Philosopher(threading.Thread):
    def __init__(self, left_chopstick, right_chopstick):
        threading.Thread.__init__(self)
        self.left_chopstick = left_chopstick
        self.right_chopstick = right_chopstick

    def run(self):
        while True:
            # 奇数号哲学家先拿左边的筷子
            if threading.current_thread().name[-1] == '1' or threading.current_thread().name[-1] == '3':
                self.left_chopstick.acquire()
                self.right_chopstick.acquire()
            # 偶数号哲学家先拿右边的筷子
            else:
                self.right_chopstick.acquire()
                self.left_chopstick.acquire()

            # 进餐
            print(f"{threading.current_thread().name} is eating...")

            # 放下右边的筷子
            self.right_chopstick.release()
            # 放下左边的筷子
            self.left_chopstick.release()

if __name__ == "__main__":
    chopsticks = [threading.Lock() for _ in range(5)]

    philosophers = []
    for i in range(5):
        philosopher = Philosopher(chopsticks[i], chopsticks[(i+1)%5])
        philosopher.name = f"Philosopher {i}"
        philosophers.append(philosopher)

    for philosopher in philosophers:
        philosopher.start()

以上两种方法都能保证不会发生死锁。方法b通过限制同时拿左边筷子的哲学家数量来避免死锁，方法c则通过奇偶号哲学家的不同拿筷子顺序来避免死锁。